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Solve the Following Equation: 4 Sin 2 X − 8 Cos X + 1 = 0 - Mathematics

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प्रश्न

Solve the following equation:

\[4 \sin^2 x - 8 \cos x + 1 = 0\]
बेरीज
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उत्तर

\[4 \sin^2 x - 8\cos x + 1 = 0\]
\[\Rightarrow 4 - 4 \cos^2 x - 8 \cos x + 1 = 0\]
\[ \Rightarrow 4 \cos^2 x + 8 \cos x - 5 = 0\]
\[ \Rightarrow 4 \cos^2 x + 10 \cos x - 2 \cos x - 5 = 0\]
\[ \Rightarrow 2 \cos x (2 \cos x + 5 ) - 1 (2 \cos x + 5) = 0\]
\[ \Rightarrow (2 \cos x - 1) (2 \cos x + 5) = 0\]
\[\Rightarrow (2 \cos x - 1) = 0\] or
\[(2 \cos x + 5) = 0\]
Now,
\[2 \cos x + 5 = 0 \Rightarrow \cos x = - \frac{5}{2}\]  (It is not possible.)

\[\therefore 2 \cos x - 1 = 0 \]

\[ \Rightarrow \cos x = \frac{1}{2} \]

\[ \Rightarrow \cos x = \cos \frac{\pi}{3} \]

\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z\]

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पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २२]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 11 Trigonometric equations
Exercise 11.1 | Q 3.4 | पृष्ठ २२

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